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The vapour pressure of benzene at 90^(@)...

The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5 g` of a solute in `58.5 g` benzene has vapour pressure `990` torr. The molecualr weight of the solute is:

A

(a) `78.2`

B

(b) `178.2`

C

( c) `206.2`

D

(d) `220`

Text Solution

Verified by Experts

The correct Answer is:
D
`(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)`
`(1020 - 990)/(990) = (5 xx 78)/(m xx 58.5)`
`:. m = 220`
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